a/ ĐKXĐ:\(x\ge0;x\ne9\)
Ta có : \(P=\frac{x\sqrt{x}-3}{x-2\sqrt{x}-3}-\frac{2\left(\sqrt{x}-3\right)}{\sqrt{x}+1}+\frac{\sqrt{x}+3}{3-\sqrt{x}}\)
\(=\frac{x\sqrt{x}-3-2\left(\sqrt{x}-3\right)^2-\left(\sqrt{x}+3\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-3\right)}\)
\(=\frac{x\sqrt{x}-3-2\left(x-6\sqrt{x}+9\right)-\left(x+4\sqrt{x}+3\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-3\right)}\)
\(=\frac{x\sqrt{x}-3-2x+12\sqrt{x}-18-x-4\sqrt{x}-3}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-3\right)}\)
\(=\frac{x\sqrt{x}-3x+8\sqrt{x}-24}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-3\right)}\)
\(=\frac{x\left(\sqrt{x}-3\right)+8\left(\sqrt{x}-3\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-3\right)}\)
\(\frac{\left(x+8\right)\left(\sqrt{x}-3\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-3\right)}=\frac{x+8}{\sqrt{x}+1}\)
Vậy P=\(\frac{x+8}{\sqrt{x}+1}\) khi \(x\ge0;x\ne9\)
b/ P\(=\frac{x+8}{\sqrt{x}+1}\)
\(=\frac{x-1}{\sqrt{x}+1}+\frac{9}{\sqrt{x}+1}=\left(\sqrt{x}-1\right)+\frac{9}{\sqrt{x}+1}\)
\(=\left(\sqrt{x}+1\right)+\frac{9}{\sqrt{x}+1}-2\)
Áp dụng BĐT Cô-si cho 2 số không âm \(\sqrt{x}+1\) và \(\frac{9}{\sqrt{x}+1}\) ta có:
P\(\ge2\sqrt{\left(\sqrt{x}+1\right).\frac{9}{\sqrt{x}+1}}-2\)
\(P\ge2.3-2=4\)
\(\Rightarrow P\ge4\)
Dấu "=" xảy ra \(\Leftrightarrow\sqrt{x}+1=\frac{9}{\sqrt{x}+1}\Leftrightarrow\left(\sqrt{x}+1\right)^2=9\Leftrightarrow\sqrt{x}+1=3\Leftrightarrow\sqrt{x}=2\Leftrightarrow x=4\)
Vậy P min =4 khi x=4
